X −→ Y Be a Continuous Map of Topological Spaces. We Say That F Is A

Home , Lifting property, Local homeomorphism

16. Maps between manifolds Definition 16.1. Let f : X −→ Y be a continuous map of topological spaces. We say that f is a local homeomorphism if for every point x ∈ X there is an open neighbourhood U of x such that V = f(U) is open and f|U : U −→ V is a homeomorphism. We say that f is an unramified cover if for every point y ∈ Y we may find an open neighbourhood V of y, such that the inverse image −1 f (V ) is a disjoint union of open subsets U such that f|U : U −→ V is a homeomorphism. Every unramified cover is a local homeomorphism. However: Example 16.2. let Y = C and X = C∗ = C−{0}. Then the inclusion of X into Y is a local homeomorphism but it is not an unramified cover. We will need some basic theory of covering spaces. Definition 16.3. Let f : X −→ Y be a continuous map of topological spaces. Let Z be any topological space and let g : Z −→ Y be any continuous map. A lift of g to X is a continuous map h: Z −→ X such that f ◦ h = g.

X - f ? Z - Y In general lifts need neither exist nor are they even unique: Example 16.4. Let f : X = C − {0} −→ C = Y be the natural inclusion. Then we cannot lift the identity map g : Y −→ Y to X. Example 16.5. Let X and Y be the spaces constructed in (13.2.2). Then there is a natural continuous map f : Y −→ X (note that we have switched X and Y , both in terms of (13.2.2) and (16.1)). However the identity map g : X −→ X is a continuous map which can be lifted in two different ways. However non-uniqueness is quite pathological: Lemma 16.6. Let f : X −→ Y be a local homeomorphism of topological spaces. Let Z be a connected topological space and let g : Z −→ Y be a continuous map. Let hi : Z −→ X, i = 1, 2 be any two maps which lift g to X. If X is Hausdorff and there is a point z0 ∈ Z such that h1(z0) = h2(z0) then h1 = h2. 1 Proof. Let E = { z ∈ Z | h1(z) = h2(z) }, be the set of points where h1 and h2 are equal. Then z0 ∈ E so that E is non-empty by assumption. As X is Hausdorff, the diagonal ∆ ⊂ X ×X is closed. Now E is the inverse image of ∆ by the continuous map h1 × h2 : Z −→ X × X, so that E is closed as X is Hausdorff. Suppose that z ∈ Z. Let x = hi(z). Then we may find an open neighbourhood U of x and V = f(U) of y = f(x) = g(z), such that f|U : U −→ V is a homeomorphism. Since hi are continuous, Wi = −1 hi (U) is open. Let W = W1 ∩ W2. As

f ◦ h1 = g = f ◦ h2, and f|U is injective, W ⊂ E. But then E is open and closed, so that E = Z as Z is connected. Lemma 16.7. Let f : X −→ Y be an unramified cover of manifolds. Let γ : [0, 1] −→ Y be a path and suppose x ∈ X is point of X such that f(x) = y = γ(0). Then γ has a unique lift to a map ψ : [0, 1] −→ X, such that ψ(0) = x. Proof. For every point y of the image of γ, we may find a connected −1 open neighbourhood V = Vy ⊂ Y of y such that f (V ) is a disjoint union of open subsets all of which are homeomorphic to V via f. Since the image of γ is compact, we may cover the image by finitely many of these open subsets V1,V2,...,Vk. Let Wj ⊂ [0, 1] be the inverse image of Vj, and let [ Ij = Wi. i≤j

Then Ij is an open subset of [0, 1] which is an interval containing 0. We define ψj : Ij −→ X by induction on j. Suppose that we have defined 0 ψk, k ≤ j. Pick t ∈ Wj ∩ Wj ⊂ Ii ∩ Wj+1. Let x = ψj(t) ∈ X (if j = 0 then we take x0 = x). Pick U ⊂ X containing x0 such that U −1 is a connected component of f (Vj+1) which maps homeomorphically down to Vj+1. Then we may clearly extend ψj to ψj+1. Uniqueness follows from (16.6). Definition 16.8. Let f : X −→ Y be a local homeomorphism. We say that f has the lifting property if given any continuous map γ : [0, 1] −→ Y and a point x ∈ X such that f(x) = y = γ(0), then we may lift γ to ψ : [0, 1] −→ X such that ψ(0) = x. 2 Theorem 16.9. Let f : X −→ Y be a local homeomorphism of manifolds. Then f has the lifting property if and only if f is an unramified cover. We have already shown one direction of (16.9). To prove the other direction we need the following basic result: Theorem 16.10. Let f : X −→ Y be a local homeomorphism of manifolds. Suppose that we have a homotopy G: [0, 1]×[0, 1] −→ Y between γ0 and γ1. Suppose also that we can lift γs : [0, 1] −→ Y , defined by γs(t) = γ(s, t) to ψs : [0, 1] −→ X. If the function σ : [0, 1] −→ X given by σ(s) = ψs(0) is continuous then we can lift G to a continuous map H : [0, 1] × [0, 1] −→ X. Proof. The definition of H : [0, 1] × [0, 1] −→ X, 2 is clear; given (s, t) ∈ [0, 1] , H(s, t) = ψs(t). The only thing we need to check is that H is continuous. Note that σ(s) = H(s, 0). Given (s, t) ∈ [0, 1]2, let x = H(s, t). Since f is a local homeomorphism, we may find an open subset U = Us,t such that V = f(U) is −1 open and f|U : U −→ V is a homeomorphism. Let W = G (V ). Then W = Ws,t is an open neighbourhood of (s, t), such that H(s, t) ∈ U and G(W ) = V . By compactness of [0, 1]2, we may find a finite subcover. It follows that we may subdivide [0, 1]2 into finitely many squares in such a way that each square is mapped by G into an open subset V of Y such that there is an open subset U of X with the property that f|U : U −→ V is a homeomorphism and there is a point (s, t) belonging to the square such that H(s, t) ∈ U. Suppose that the subdivision is given by k, so that we subdivide each [0, 1] into the intervals [i/k, (i + 1)/k]. By an obvious induction, it suffices to prove that

H|[i/k,(i+1)/k]×[j/k,(j+1)/k], is continuous, given that σ(t) = H(i/k, t) is continuous. Replacing the square [i/k, (i + 1)/k] × [j/k, (j + 1)/k], by [0, 1]2 and Y by V , we are reduced to proving that H is continuous, given that there is an open subset U of X which maps homeomorphi- 2 cally down to Y and such that there is a single point (s0, t0) ∈ [0, 1] such that H(s0, t0) ∈ U.

Since f|U : U −→ Y is a homeomorphism, we may lift γs0 to a map ψ0 : [0, 1] −→ U. By uniqueness it follows that H(s0, t) ∈ U for all t ∈ [0, 1]. By uniqueness of the lift of γ, where γ(t) = G(0, t), H(0, t) ∈ U 3 for all t ∈ [0, 1]. Finally by uniqueness of the lift of γs(t), it follows 2 −1 that H(s, t) ∈ U for all (s, t) ∈ [0, 1] . But then G = (f|U ) ◦ H is certainly continuous. To prove (16.9), we start with a seemingly special case: Proposition 16.11. Let f : X −→ Y be a local homeomorphism of connected manifolds. If f has the lifting property and Y is simply connected then f is an isomorphism. Proof. Pick a ∈ X and let b = f(a) ∈ Y . Pick y ∈ Y . Since Y is a connected manifold it is path connected. Pick a path γ : [0, 1] −→ Y such that b = γ(0) and y = γ(1). Let ψ : [0, 1] −→ X be a lifting of γ and let x = ψ(1). Then f(x) = y. Thus f is surjective. Suppose that x0, x1 ∈ X, with y = f(x0) = f(x1). Pick paths ψi : [0, 1] −→ X such that a = ψi(0) and xi = ψi(1). Let γi = f ◦ ψi. As Y is simply connected, we may pick a homotopy G between ψ0 and ψ1 ﬁxing y. By (16.10) we may lift G to a continuous function H, such that H(0, 0) = x. Since the function ψ(t) = G(0, t) is continuous, the composition f ◦ ψ is constant (equal to y) and the ﬁbre f −1(y) is discrete, it follows that G(0, t) = x is constant. But then x1 = G(0, 1) = G(0, 0) = x0. Thus f is injective. As f is a local homeomorphism, it follows that f is a homeomorphism. Proof of (16.9). Suppose that f has the lifting property. Pick y ∈ Y and let V be a simply connected neighbourhood of y. Let U be a connected component of the inverse image. Then f|U : U −→ V has the lifting property and by (16.11) it is a homeomorphism.